25 march 2013birkbeck college, u. london1 introduction to programming lecturer: steve maybank...

25 March 2013 Birkbeck College, U. London 1

Introduction to Programming

Lecturer: Steve Maybank

Department of Computer Science and Information Systems

[email protected] 2013

Week 11: Examples of Algorithms

JavaLab 9, Ex. 1 (1)

Write a method

public static int[] reverseArray(int[] data)

such that reverseArray returns the reverse of the array data. Call reverseArray from main.


JavaLab 9, Ex. 1 (2)import java.util.Arrays;/** * This program reverses the order of the elements in an

array * @author S.J. Maybank * @version 25 March 2013 */public class ReverseArray{ // main defined here // reverseArray defined here}


JavaLab 9, Ex. 1 (3)public static void main(String[] args){ int[] data1 = {1, 2, 3, 4, 5}; int[] data2 = {}; int[] data3 = {1, 2, 3, 4}; int[] data1R = reverseArray(data1); int[] data2R = reverseArray(data2); int[] data3R = reverseArray(data3); System.out.println("Original array: "+Arrays.toString(data1)); System.out.println("Reversed array:

"+Arrays.toString(data1R)); // data2, dataR and data3, data3R printed out similarly}


JavaLab 9, Ex. 1 (4)public static int[] reverseArray(int[] data){ int[] dataReversed = new int[data.length]; for(int i = 0; i < data.length; i++) { dataReversed[data.length-1-i] = data[i]; } return dataReversed;}


JavaLab 9, Ex. 2 (1)/** * A program to apply certain methods to an array of integers. * @author S.J. Maybank * @version 25 March 2013 */public class ArrayMethods{

// main // printArray

// productElements// numberNegativeElements

}


JavaLab 9, Ex. 2 (2)

public static void main(String[] args){ int[] data = {1, 2, 3, -1, -3}; printArray(data); System.out.println("Product elements:

"+productElements(data)); System.out.print("Number of elements strictly less than 0:

"); System.out.println(""+numberNegativeElements(data));}


JavaLab 8, Ex. 2 (3)public static void printArray(int[] data){ for(int i = 0; i < data.length; i++) { System.out.print(data[i]); if (i < data.length-1) { System.out.print(" "); } } System.out.println();}


JavaLab 9, Ex. 2 (4) public static int productElements(int[] data)

{ int product = 1; for(int i = 0; i < data.length; i++) { product *= data[i]; } return product; }


JavaLab 9. Ex. 2 (5)public static int numberNegativeElements(int[] data){ int n = 0; for(int i = 0; i < data.length; i++) { if(data[i] < 0) { ++n; } } return n;}


Overview

Test to see if two appointments overlap (JFE, R3.11)

Linear search of an array (JFE, Section 6.3.5)

Binary search of an array (JFE, end of Section 6.3)


Appointments


timea1 a2 a3 a4

Non-overlapping appointments: [a1, a2] and [a3, a4]Overlapping appointments:

[a1, a3] and [a2, a4][a1, a4] and [a2, a3]

How to decide when two appointments overlap?

Overlapping Appointments Let the appointments be [s1, e1] and [s2, e2].

Let s be the latest start and let e be the earliest end.

If s > e, then one appointment begins after the other has finished.

Conversely, let t be any time such that s ≤ t ≤ e. The time t is in [s1, e1] because

s1 s ≤ t ≤ e e1 The time t is in [s2, e2] because

s2 s ≤ t ≤ e e2


Pseudo Code for overLappingAppointments

Inputs: times s1, e1 and s2, e2 for two appointments.

Output: true if the appointments overlap and false otherwise

if (s1 > s2)s = s1 else s = s2

if (e1 < e2)e = e1 else e = e2

if (s < e) return true else return false


The Method overLappingAppointments

public static boolean overLappingAppointments(int s1, int e1, int s2, int e2)

{int s, e;if(s1 > s2){s = s1;} else {s = s2;}if(e1 < e2){e = e1;} else {e = e2;}return s < e;

}


Inputs and Output for Linear Search

Inputs: 1D integer array data and an integer e.

Output: if data contains e then an integer pos such that

data[pos] == e,otherwise –1.


Pseudo Code for Linear Search

1. Step through the valid indices pos for the array data,

if data[pos] == e, then return pos.

2. If all the valid indices have been checked without finding e, then return -1


The Method linearSearchpublic static int linearSearch(int[] data, int e){

int pos = 0;while (pos < data.length){

if(data[pos] == e){return pos;}++pos;

}return –1;

}


The Method linearSearch2public static int linearSearch2(int[] data, int e){

int pos = data.length-1;while (pos>=0){

if(data[pos] == e){return pos;}pos--;

}return –1;

}


Inputs and Output for Binary Search

Inputs: 1D sorted integer array data and an integer e.

Output: if data contains e then an integer pos such that

data[pos] == e,otherwise –1.


Strategy for Binary Search


1 3 7 8 11 12 20

low high

Mark out a section of the array data using indices low, highFind an index i between low and highCompare data[i] with e and update low, high accordingly

i

Pseudo Code for Binary Search

Set low equal to the least index for data.Set high equal to the largest index for data.while (low <= high){

Find an index pos between low and high.if (data[pos] == e) then return pos.if (data[pos] < e) then high = pos-1.if (data[pos] > e) then low = pos+1.

}return –1.


The Method binarySearchpublic static int binarySearch(int[] data, int e){

int low = 0, high = data.length-1, pos = 0;while(low <= high){

pos = (low+high)/2;if (data[pos] == e){return pos;}if (data[pos] < e){low = pos+1;} // look in second halfelse{high = pos-1;} // look in first half

}return –1;

}


Example of Binary Search Inputs: data = {1, 3, 5, 7, 9, 11}, e = 6.low = 0, high = 5, pos = 2.data[pos] < e, thus low = pos+1 = 3.low = 3, high = 5, pos = 4.data[pos] > e, thus high = pos-1 = 3.low = 3, high = 3, pos = 3.data[pos] > e, thus high = pos-1 = 2.low = 3, high =2(low <= high) == false, search terminates.


25 march 2013birkbeck college, u. london1 introduction to programming lecturer: steve maybank...

Documents

birkbeck college

int pos

int s2

int e2

int e1

pos data

int datareversed

e1 e2 e